Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a cycle in the linked list, where tail connects to the second node. 

Example 2:

Input: head = [1,2], pos = 0 Output: true Explanation: There is a cycle in the linked list, where tail connects to the first node. 

Example 3:

Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list. 

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

这道题是快慢指针的经典应用。只需要设两个指针，一个每次走一步的慢指针和一个每次走两步的快指针，如果链表里有环的话，两个指针最终肯定会相遇。实在是太巧妙了，要是我肯定想不出来。代码如下：

C++ 解法：

class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) return true;
        }
        return false;
    }
};

Java 解法：

public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) return true;
        }
        return false;
    }
}

类似题目：

Linked List Cycle II

Happy Number

参考资料：

https://leetcode.com/problems/linked-list-cycle/

https://leetcode.com/problems/linked-list-cycle/discuss/44489/O(1)-Space-Solution